Solution manual for a course in probability by weiss

CHAPTER ONE

Instructor’s
Solutions Manual
Anna Amirdjanova
University of Michigan, Ann Arbor

Neil A. Weiss
Arizona State University

FOR
A Course In

Probability
Neil A. Weiss
Arizona State University

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Compositor: Anna Amirdjanova and Neil A. Weiss
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Chapter 1

Probability Basics
1.1 From Percentages to Probabilities
Basic Exercises
1.1
a) The population under consideration consists of the five senators specified in the problem statement,
namely, Graham, Baucus, Conrad, Murkowski, and Kyl.

b) Because three of the five senators are Democrats and each senator is equally likely to be the one
selected, the probability is 3/5, or 0.6, that the chosen senator is a Democrat.
1.2 We represent each possible subcommittee as a subset of two of the the five senators. Also, for
convenience, we use the first letter of each senator’s last name to represent the senator.
a) The 10 possible subcommittees are as follows: {G, B}, {G, C}, {G, M}, {G, K}, {B, C}, {B, M}, {B, K},
{C, M}, {C, K}, {M, K}.
b) Of the 10 possible subcommittees, exactly three consist of two Democrats, namely, {G, B}, {G, C},
and {B, C}. Hence, the probability that the committee will consist of two Democrats is 3/10, or 0.3.
c) Of the 10 possible subcommittees, exactly one consists of two Republicans, namely, {M, K}. Hence,
the probability that the committee will consist of two Republicans is 1/10, or 0.1.
d) Of the 10 possible subcommittees, exactly six consist of one Democrat and one Republican, namely,
{G, M}, {G, K}, {B, M}, {B, K}, {C, M}, and {C, K}. Hence, the probability that the committee will
consist of one Democrat and one Republican is 6/10, or 0.6.
1.3 We represent each possible chair/vice-chair possibility as an ordered pair of two of the five senators,
the first entry being the chair and the second the vice-chair. Also, for convenience, we use the first letter
of each senator’s last name to represent the senator.
a) The 20 possibilities are (G, B), (B, G), (G, C), (C, G), (G, M), (M, G), (G, K), (K, G), (B, C),
(C, B), (B, M), (M, B), (B, K), (K, B), (C, M), (M, C), (C, K), (K, C), (M, K), (K, M).
b) Equally probable. Indeed, in the scenario described in Exercise 1.1, the probability that Graham will
be chosen as chair is 1/5, or 0.2. In the current scenario, of the 20 possibilities, exactly four have Graham
as chair, namely, (G, B), (G, C), (G, M), and (G, K); thus, the probability that Graham will be chosen
as chair is 4/20, or 0.2.
c) Of the 20 possibilities, exactly 12 have a Democrat as chair, namely, (G, B), (B, G), (G, C), (C, G),
(G, M), (G, K), (B, C), (C, B), (B, M), (B, K), (C, M), and (C, K). Hence, the probability that the
senator chosen to be chair is a Democrat is 12/20, or 0.6.
1.4 We work in number of thousands of housing units. From the table, we see that the total number of
housing units is
471 + 1,470 + · · · + 15,647 = 112,356.

1-1

1-2

CHAPTER 1 Probability Basics

a) The number of housing units with (exactly) four rooms is 23,468. Hence, the probability that the
housing unit selected has four rooms is 23,468/112,356 ≈ 0.209.
b) The number of housing units with more than four rooms is
24,476 + 21,327 + 13,782 + 15,647 = 75,232.
Thus, the probability that the housing unit selected has more than four rooms is 75,232/112,356 ≈ 0.670.
c) The number of housing units with either one or two rooms is 471 + 1,470 = 1,941. Hence, the
probability that the housing unit selected has either one or two rooms is 1,941/112,356 ≈ 0.0173.
d) None of the housing units have fewer than one room. Hence, the probability that the housing unit
selected has fewer than one room is 0/112,356 = 0.
e) All of the housing units have one or more rooms. Hence, the probability that the housing unit selected
has one or more rooms is 112,356/112,356 = 1.
f) The population under consideration consists of all U.S. housing units.
1.5 From the table, we see that the total number of murder cases during the year in question in which
the person murdered was between 20 and 59 years old, inclusive, is
2,916 + 2,175 + · · · + 372 = 11,527.
a) The number of these murder cases in which the person murdered was between 40 and 44 years old,
inclusive, is 1213. Hence, the probability that the murder victim of the case selected was between 40 and
44 years old, inclusive, is 1,213/11,527 ≈ 0.105.
b) We see that the number of these murder cases in which the person murdered was 25 years old or older
is 11,527 − 2,916 = 8,611. Consequently, the probability that the murder victim of the case selected
was 25 years old or older is 8,611/11,527 ≈ 0.747.
c) The number of these murder cases in which the person murdered was between 45 and 59 years old,
inclusive, is
888 + 540 + 372 = 1800.

Hence, the probability that the murder victim of the case selected was between 45 and 59 years old,
inclusive, is 1,800/11,527 ≈ 0.156.
d) The number of these murder cases in which the person murdered was either under 30 or over 54 is
2916 + 2175 + 372 = 5463.
Consequently, the probability that the murder victim of the case selected was either under 30 or over 54
is 5,463/11,527 ≈ 0.474.
e) The population under consideration consists of all murder cases during the year in question in which
the person murdered was between 20 and 59 years old, inclusive.
1.6 Answers will vary, depending on the students in your class. Let us, however, denote by ℓ1 and ℓ2 the
number of left-handed females and left-handed males in your class, respectively, and by n1 and n2 the
number of females and males in your class, respectively. Note that the total number of students in your
class is represented by n1 + n2 .
a) The number of females in your class is n1 . Hence, the probability that a randomly selected student in
your class is female is n1 /(n1 + n2 ).
b) The number of left-handed students in your class is ℓ1 + ℓ2 . Hence, the probability that a randomly
selected student in your class is left-handed is (ℓ1 + ℓ2 )/(n1 + n2 ).
c) The number of left-handed females in your class is ℓ1 . Hence, the probability that a randomly selected
student in your class is a left-handed female is ℓ1 /(n1 + n2 ).


1.1

From Percentages to Probabilities

1-3

d) A student is neither female nor left-handed if and only if the student is a right-handed male. The
number of right-handed males in your class is the number of males who are not left-handed, which
is n2 − ℓ2 . Hence, the probability that a randomly selected student in your class is neither female nor
left-handed is (n2 − ℓ2 )/(n1 + n2 ).

1.7 Because we are selecting at random from a finite population (the students at the university in question),
probabilities are the same as percentages.
a) We know that 62% of the students are bilingual and that 80% of those students speak Spanish.
Now, 80% of 62% is 49.6%. Hence, 49.6% of the students speak Spanish, so that 50.4% don’t. Thus,
the probability that a randomly selected student at this university doesn’t speak Spanish is 0.504.
b) From part (a), 49.6% of the students speak Spanish and we know that 10% of the Spanish-speaking
students also speak French. Now, 10% of 49.6% is 4.96%. Hence, 4.96% of the students speak both
Spanish and French. Consequently, the probability that a randomly selected student at this university
speaks both Spanish and French is 0.0496.
1.8 Answers will vary. However, to illustrate, we tossed a balanced coin 20 times and obtained the
following table. Here E denotes the event that, on any particular toss, the coin comes up a head.
Toss
n

Outcome
(H or T)

Number of heads
n(E )

Proportion of heads
n(E )/n

1
2
3
4
5
6
7

8
9
10
11
12
13
14
15
16
17
18
19
20

T
H
H
T
H
T
T
T
H
H
T
T
T
H
H
H

H
T
T
T

0
1
2
2
3
3
3
3
4
5
5
5
5
6
7
8
9
9
9
9

0.000
0.500
0.667
0.500

0.600
0.500
0.429
0.375
0.444
0.500
0.455
0.417
0.385
0.429
0.467
0.500
0.529
0.500
0.474
0.450

a) See the first five rows of the preceding table.
b) Referring to the frequentist interpretation of probability on page 5 and the fifth row of the preceding
table, we estimate, based on the first five tosses of the coin, that the probability of a head is 3/5 = 0.6.
c) Referring to the frequentist interpretation of probability and the 10th row of the preceding table, we
estimate, based on the first 10 tosses of the coin, that the probability of a head is 5/10 = 0.5.
d) Referring to the frequentist interpretation of probability and the 20th row of the preceding table, we
estimate, based on the first 20 tosses of the coin, that the probability of a head is 9/20 = 0.45.
e) We see from parts (b)–(d) that, based on the frequentist interpretation of probability, our estimate of
the probability of a head will fluctuate. This phenomenon will be the case regardless of how many times
we toss the coin.


CHAPTER 1 Probability Basics

1-4

1.9 Note that 2004 was a leap year, so it had 366 days. Now, consider the random experiment of
selecting one 2004 U.S. state governor at random, and let E denote the event that the governor chosen
is a Republican. We know that P (E) = 28/50 = 0.56. Thus, from the frequentist interpretation of
probability, if we independently repeat the random experiment n times, then the proportion of times that
event E occurs will approximately equal 0.56. Consequently, in 366 repetitions of the random experiment,
we would expect event E to occur roughly 0.56 · 366 = 204.96 times. In other words, if on each of the
366 days of 2004, one U.S. state governor was randomly selected to read the invocation on a popular
radio program, then we would expect a Republican to be chosen on approximately 205 of those days.
1.10 In each case, we refer to the frequentist interpretation of probability on page 5.
a) Approximately 31.4% of human gestation periods exceed 9 months.
b) The favorite finishes in the money in roughly two-thirds of horse races.
c) About 40% of traffic fatalities involve an intoxicated or alcohol-impaired driver or nonoccupant.
1.11 Let E be an event and let P (E) denote its probability. For n independent repetitions of the random
experiment, let n(E) denote the number of times that event E occurs. Then, it follows from Relation (1.1)
on page 5 that
n(E) ≈ nP (E),
for large n.
(∗)
a) The random experiment consists of observing the duration of a human gestation period and event E
is that the gestation period exceeds 9 months. We know that P (E) = 0.314. Here n = 4000 so that
from Relation (∗),
n(E) ≈ 4000 · 0.314 = 1256.
Hence, of 4000 human gestation periods, roughly 1256 will exceed 9 months.
b) The random experiment consists of observing a horse race and event E is that the favorite finishes in
the money. We know that P (E) = 2/3. Here n = 500 so that from Relation (∗),
2
1000

=
≈ 333.3.
3
3
Therefore, in 500 horse races, the favorite will finish in the money about 333 times.
c) The random experiment consists of observing a traffic fatality (or reading a traffic-fatality report) and
event E is that the traffic fatality involves an intoxicated or alcohol-impaired driver or nonoccupant. We
know that P (E) = 0.40. Here n = 389 so that from Relation (∗),
n(E) ≈ 500 ·

n(E) ≈ 389 · 0.40 = 155.6.
Therefore, in 389 traffic fatalities, approximately 156 will involve an intoxicated or alcohol-impaired
driver or nonoccupant.

Advanced Exercises
1.12 The probability that the ball lands on a red number is 18/38, so that the odds against that event are
1 − 18/38 to 18/38

or

20/38 to 18/38

or

20 to 18

or

10 to 9.

Hence, to make the bet fair, the house should pay 10 to 9 for a bet on red; that is, for a winning bet on
red, the house should pay the gambler $10 for each $9 bet.
1.13
a) As the odds against Fusaichi Pegasus were 3 to 5, the probability that Fusaichi Pegasus would win
the race is 5/(5 + 3) = 0.625.
b) Because the odds against Red Bullet were 9 to 2, the probability that Red Bullet would win the
race is 2/(2 + 9) ≈ 0.182.


1.2

Set Theory

1-5

1.2 Set Theory
Note: For convenience, we use “iff” to represent the phrase “if and only if.”

Basic Exercises
1.14 To avoid trivialities, we assume that there are at least two sets in the collection. Hence, let A1 , A2 , . . .
be pairwise disjoint sets. Then, in particular, we have A1 ∩ A2 = ∅. Consequently,
An ⊂ A1 ∩ A2 = ∅.
n

As the empty set is a subset of every set, we conclude that n An = ∅. Therefore, if the sets in a collection
are pairwise disjoint, then the intersection of all the sets in the collection must be empty.
1.15 Answers will vary, but here is one possibility:

1.16 Answers will vary, but here is one possibility:

1.17 Answers will vary, but the subsets of R2 portrayed in the solution to Exercise 1.16 provide one
possibility. For another possibility, let U = {a, b, c, d, e, f } and consider the following four subsets
of U : {a, b, e}, {b, c, f }, {c, d, e}, {a, d, f }. We note that each pairwise intersection is nonempty and,
hence, in particular, the four sets are not pairwise disjoint; however, every three sets have an empty
intersection.
1.18
a) We note that A1 ⊃ A2 ⊃ · · · . Therefore,
4

4

An = A4 = [0, 1/4]
n=1

An = A1 = [0, 1].

and
n=1


CHAPTER 1 Probability Basics

1-6

b) We have x ∈ ∞
n=1 An iff x ∈ An for all n ∈ N iff x ∈ [0, 1/n] for all n ∈ N iff 0 ≤ x ≤ 1/n for
all n ∈ N iff x = 0. Hence,
∞

An = {0}.

n=1

As A1 ⊃ A2 ⊃ · · · , we have
∞

An = A1 = [0, 1].
n=1

1.19 In each part, let An denote the set in the union or intersection. It will be useful to note the following
facts, which hold for all real numbers c.
●

x < c iff x ≤ c − 1/n for some n ∈ N

●

x > c iff x ≥ c + 1/n for some n ∈ N

●

x ≤ c iff x < c + 1/n for all n ∈ N

●

x ≥ c iff x > c − 1/n for all n ∈ N

a) We have x ∈ ∞
n=1 An iff x ∈ An for some n ∈ N iff 1 + 1/n ≤ x ≤ 2 − 1/n for some n ∈ N
iff 1 < x < 2. Hence,
∞

[1 + 1/n, 2 − 1/n] = (1, 2).
n=1

b) We have x ∈
Hence,

∞
n=1 An iff x

∈ An for some n ∈ N iff 1 ≤ x ≤ 2 − 1/n for some n ∈ N iff 1 ≤ x < 2.
∞

[1, 2 − 1/n] = [1, 2).
n=1

c) We note that x ∈
iff 1 ≤ x ≤ 2. Hence,

∞
n=1 An

iff x ∈ An for all n ∈ N iff 1 − 1/n < x < 2 + 1/n for all n ∈ N
∞

(1 − 1/n, 2 + 1/n) = [1, 2].
n=1

d) We note that x ∈
iff 3 ≤ x ≤ 3. Hence,

∞
n=1 An

iff x ∈ An for all n ∈ N iff 3 − 1/n < x < 3 + 1/n for all n ∈ N
∞

(3 − 1/n, 3 + 1/n) = {3}.
n=1

e) We have x ∈ ∞
n=1 An iff x ∈ An for all n ∈ N iff x > n for all n ∈ N . However, there are no real
numbers x that satisfy x > n for all n ∈ N . Hence,
∞

(n, ∞) = ∅.
n=1

f) We have x ∈ ∞
n=1 An iff x ∈ An for all n ∈ N iff 5 − 1/n < x < 5 for all n ∈ N iff 5 ≤ x < 5,
which is impossible. Hence,
∞

(5 − 1/n, 5) = ∅.
n=1


1.2

g) We have x ∈

Hence,

∞
n=1 An

Set Theory

1-7

iff x ∈ An for all n ∈ N iff 5 − 1/n < x < 6 for all n ∈ N iff 5 ≤ x < 6.
∞

(5 − 1/n, 6) = [5, 6).
n=1

1.20
a) From Definition 1.5 on page 16, we see that
{1, 2, 3} × {3, 4, 5} = {(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5)}.
Thus, the members of {1, 2, 3} × {3, 4, 5} are the nine ordered pairs inside the curly braces on the right
of the preceding display.
b) We determined {1, 2, 3} × {3, 4, 5} in part (a). Proceeding similarly, we find that
{3, 4, 5} × {1, 2, 3} = {(3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3)}.
Noting that the only element common to {1, 2, 3} × {3, 4, 5} and {3, 4, 5} × {1, 2, 3} is (3, 3), we see that
the members of the union of those two sets are (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2),
(3, 3), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), and (5, 3).
c) We note that U has 25 members, namely, all ordered pairs of the form (x, y) where x, y ∈ {1, 2, 3, 4, 5}.
Hence, A consists of the eight elements in U not listed in the solution to part (b), namely, (1, 1), (1, 2),
(2, 1), (2, 2), (4, 4), (4, 5), (5, 4), and (5, 5).
d) We have
{(1, 1), (1, 2), (2, 1), (2, 2)} = {1, 2} × {1, 2}

and

{(4, 4), (4, 5), (5, 4), (5, 5)} = {4, 5} × {4, 5}.

Therefore, from part (c),
A = {1, 2} × {1, 2} ∪ {4, 5} × {4, 5} .
1.21 Refer to Definition 1.5 on page 16.
a) We have
{0, 1}3 = {0, 1} × {0, 1} × {0, 1}
= {(0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 0), (1, 0, 1), (1, 1, 0), (1, 1, 1)}.
Hence, the members of {0, 1}3 are the eight ordered triplets inside the curly braces on the second line of
the preceding display.
b) We have
{0, 1} × {0, 1} × {1, 2} = {(0, 0, 1), (0, 0, 2), (0, 1, 1), (0, 1, 2), (1, 0, 1), (1, 0, 2), (1, 1, 1), (1, 1, 2)}.
Thus, the members of {0, 1} × {0, 1} × {1, 2} are the eight ordered triplets inside the curly braces on the
right of the preceding display.
c) We have
{a, b} ∪ {c, d, e} × {f, g, h} = {a, b, c, d, e} × {f, g, h}
= {(a, f ), (a, g), (a, h), (b, f ), (b, g), (b, h), (c, f ), (c, g),
(c, h), (d, f ), (d, g), (d, h), (e, f ), (e, g), (e, h)}.
Hence, the members of {a, b} ∪ {c, d, e} × {f, g, h} are the 15 ordered pairs shown in the preceding
display.


CHAPTER 1 Probability Basics

1-8

d) We have

{a, b} × {f, g, h} = {(a, f ), (a, g), (a, h), (b, f ), (b, g), (b, h)}
and
{c, d, e} × {f, g, h} = {(c, f ), (c, g), (c, h), (d, f ), (d, g), (d, h), (e, f ), (e, g), (e, h)}.
Therefore,
{a, b} × {f, g, h} ∪ {c, d, e} × {f, g, h}
= {(a, f ), (a, g), (a, h), (b, f ), (b, g), (b, h), (c, f ), (c, g), (c, h), (d, f ), (d, g), (d, h),
(e, f ), (e, g), (e, h)}.
Hence, the members of {a, b} × {f, g, h} ∪ {c, d, e} × {f, g, h} are the 15 ordered pairs shown in the
preceding display. Note that the answer here is identical to that in part (c). This fact is due to the identity
(A ∪ B) × C = (A × C) ∪ (B × C).
1.22 We first establish the identity
(A × B) ∩ (C × D) = (A ∩ C) × (B ∩ D).
Indeed, (x, y) ∈ (A × B) ∩ (C × D) iff (x, y) ∈ A × B and (x, y) ∈ C × D iff x ∈ A and y ∈ B and
x ∈ C and y ∈ D iff x ∈ A ∩ C and y ∈ B ∩ D iff (x, y) ∈ (A ∩ C) × (B ∩ D). Applying the preceding
identity, we get
[0, 2] × [0, 2] ∩ [1, 3] × [1, 3] = [0, 2] ∩ ([1, 3] × [0, 2] ∩ [1, 3] = [1, 2] × [1, 2].
1.23 Answers will vary. One possibility is to take In = (0, 1/n) for n ∈ N . Clearly, we have In ⊂ [0, 1]
for all n ∈ N . Also, for j, k ∈ N ,
Ij ∩ Ik = (0, 1/j ) ∩ (0, 1/k) = 0, 1/ max{j, k} = ∅.
Moreover, x ∈ ∞
n=1 In iff x ∈ In for all n ∈ N iff 0 < x < 1/n for all n ∈ N iff 0 < x ≤ 0, which is
impossible. Hence, ∞
n=1 In = ∅.
1.24 Answers will vary. One possibility is to take In = [0, 1/n) for n ∈ N . Indeed, x ∈ ∞
n=1 In
iff x ∈ In for all n ∈ N iff 0 ≤ x < 1/n for all n ∈ N iff 0 ≤ x ≤ 0 iff x = 0. Hence, ∞
I
=
{0}.
n=1 n

1.25 Answers will vary. One possibility is to take In = (n − 1, n] for n ∈ Z. Clearly, these intervals
constitute a countably infinite collection of intervals of R, each of length 1. Also, suppose that j, k ∈ Z
with j = k, say, j < k. If x ∈ Ij , then x ≤ j ≤ k − 1 and, hence, x ∈ Ik . Consequently, we see
that Ij ∩ Ik = ∅. Now, let x ∈ R and set n = min{j ∈ Z : x ≤ j }. Then n − 1 < x ≤ n so that x ∈ In .
Therefore, we have shown that
∞

R=

In =
n∈Z

(n − 1, n].
n=−∞

Theory Exercises
1.26
a) By observing that the last Venn diagrams in the following two rows are identical, we conclude
that (A ∩ B)c = Ac ∪ B c .


1.2

Set Theory

1-9

/ A ∩ B so that x ∈
/ A or x ∈
/ B. But then x ∈ Ac or x ∈ B c ,

b) Suppose that x ∈ (A ∩ B)c . Then x ∈
c
c
c
c
c
which means x ∈ A ∪ B . Consequently, (A ∩ B) ⊂ A ∪ B . Conversely, suppose that x ∈ Ac ∪ B c .
Then x ∈ Ac or x ∈ B c so that x ∈
/ A or x ∈
/ B. But then x ∈
/ A ∩ B, which means x ∈ (A ∩ B)c . Conc
c
c
sequently, A ∪ B ⊂ (A ∩ B) . We have shown that (A ∩ B)c ⊂ Ac ∪ B c and Ac ∪ B c ⊂ (A ∩ B)c .
Therefore, (A ∩ B)c = Ac ∪ B c .
c
c) We first note that E c = E for any set E. Now we apply part (a) of De Morgan’s laws to the sets Ac
and B c to obtain
(A ∩ B)c =

Ac

c

∩ Bc

c c

=

Ac ∪ B c

c c

= Ac ∪ B c .

Hence, we have verified part (b) of De Morgan’s laws.
1.27
a) We have the following Venn diagrams:

As the last Venn diagrams in the two preceding rows are identical, we conclude that
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).


CHAPTER 1 Probability Basics

1-10

We have the following Venn diagrams:

As the last Venn diagrams in the two preceding rows are identical, we conclude that
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
b) For part (a) of the distributive laws, we have x ∈ A ∩ (B ∪ C) iff x ∈ A and either x ∈ B or x ∈ C iff
either x ∈ A and x ∈ B or x ∈ A and x ∈ C iff either x ∈ A ∩ B or x ∈ A ∩ C iff x ∈ (A ∩ B) ∪ (A ∩ C).
Hence,
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).
We can use a similar argument to establish part (b) of the distributive laws. Alternatively, we can apply
De Morgan’s laws and part (a) of the distributive laws (which we just verified), as follows:
A ∪ (B ∩ C) = Ac ∩ (B ∩ C)c
=

Ac ∩ B c

c

∩

c

= Ac ∩ B c ∪ C c
Ac ∩ C c

c

=

Ac

c

=
c

c

Ac ∩ B c ∪ Ac ∩ C c

∪ Bc

c

∩

Ac

c

∪ Cc

c

= (A ∪ B) ∩ (A ∪ C).
1.28
a) We observe that the two commutative laws [i.e., parts (a) and (b) of Proposition 1.3] are clear from
the following two Venn diagrams:


1.2

Set Theory

1-11

For the first associative law [i.e., part (c) of Proposition 1.3], we have the following Venn diagrams:

As the last Venn diagrams in the two preceding rows are identical, we conclude that
A ∩ (B ∩ C) = (A ∩ B) ∩ C.
For the second associative law [i.e., part (d) of Proposition 1.3], we have the following Venn diagrams:

As the last Venn diagrams in the two preceding rows are identical, we conclude that

A ∪ (B ∪ C) = (A ∪ B) ∪ C.


CHAPTER 1 Probability Basics

1-12

b) First we prove part (a) of Proposition 1.3. We have x ∈ A ∩ B iff x ∈ A and x ∈ B iff x ∈ B and x ∈ A
iff x ∈ B ∩ A. Hence,
A ∩ B = B ∩ A.
We can use a similar argument to establish part (b) of Proposition 1.3. Alternatively, we can apply part (a)
of that proposition (which we just verified) and De Morgan’s laws, as follows:
A ∪ B = Ac

c

∪ Bc

c

= Ac ∩ B c

c

= B c ∩ Ac

c

= Bc

c

∪ Ac

c

= B ∪ A.

Next we prove part (c) of Proposition 1.3. We have x ∈ A ∩ (B ∩ C) iff x ∈ A, and x ∈ B and x ∈ C
iff x ∈ A and x ∈ B, and x ∈ C iff x ∈ A ∩ B and x ∈ C iff x ∈ (A ∩ B) ∩ C. Hence,
A ∩ (B ∩ C) = (A ∩ B) ∩ C.
We can use a similar argument to establish part (d) of Proposition 1.3. Alternatively, we can apply part (c)
of that proposition (which we just verified) and De Morgan’s laws, as follows:
A ∪ (B ∪ C) = Ac
=

c

∪

Bc

c

= Ac

c

∪ Bc ∩ Cc

= Ac ∩ B c

c

∪ Cc

∪ Cc

Ac ∩ B c ∩ C c

c

c

c

=

c

= Ac ∩ B c ∩ C c

Ac

c

∪ Bc

c

∪ Cc

c
c

= (A ∪ B) ∪ C.
1.29
a) By definition, A ∪ ∅ consists of all elements that are either in A or ∅. But, ∅ contains no elements.
Hence, we must have A ∪ ∅ = A.
b) If x ∈ A, then x ∈ A or x ∈ B, which means that x ∈ A ∪ B. Hence, A ⊂ A ∪ B.
c) Suppose that A = A ∪ B. Then, from part (b),
B ⊂ B ∪ A = A ∪ B = A.
Hence, B ⊂ A. Conversely, suppose that B ⊂ A. Let x ∈ A ∪ B. Then either x ∈ A or x ∈ B. However, as B ⊂ A, if x ∈ B, then x ∈ A. Hence, in either case, we have x ∈ A. Thus, we have shown
that A ∪ B ⊂ A. From part (b), A ⊂ A ∪ B. Consequently, we have A = A ∪ B.
1.30
a) By definition, A ∩ ∅ consists of all elements that are both in A and ∅. But, ∅ contains no elements.
Hence, we must have A ∩ ∅ = ∅.
b) If x ∈ A ∩ B, then x ∈ A and x ∈ B, so that, in particular, x ∈ A. Consequently, A ∩ B ⊂ A or,
equivalently, A ⊃ A ∩ B.
c) Suppose that A = A ∩ B. Then, from part (b),
B ⊃ B ∩ A = A ∩ B = A.
Hence, B ⊃ A. Conversely, suppose that B ⊃ A. Let x ∈ A. Then, as B ⊃ A, we must have x ∈ B.
Therefore, x ∈ A ∩ B. Thus, we have shown that A ⊂ A ∩ B. From part (b), A ⊃ A ∩ B. Consequently, A = A ∩ B.
1.31
a) Let x ∈ A. Now, either x ∈ B or x ∈ B c . In the former case, we then have x ∈ A ∩ B, whereas, in
the latter case, x ∈ A ∩ B c . Thus, x ∈ (A ∩ B) ∪ (A ∩ B c ). Hence, A ⊂ (A ∩ B) ∪ (A ∩ B c ). From
Exercise 1.30(b),
A⊃A∩B
and
A ⊃ A ∩ Bc.

Hence, A ⊃ (A ∩ B) ∪ (A ∩ B c ). Therefore, we have now shown that A = (A ∩ B) ∪ (A ∩ B c ).


1.2

Set Theory

1-13

b) Suppose that A ∩ B = ∅. Let x ∈ A. Then x ∈
/ B so that x ∈ B c . Hence, A ⊂ B c .
c) Suppose that A ⊂ B. Let x ∈ B c . Then x ∈
/ B so that x ∈
/ A (because if x ∈ A, then x ∈ B).
Hence, x ∈ Ac . Consequently, we have shown that B c ⊂ Ac .
c

1.32 We have x ∈
/ n An iff x ∈
/ An for some n iff x ∈ Acn for some n iff x ∈
n An iff x ∈
Hence, Proposition 1.4(a) holds. Applying that result to the sets Ac1 , Ac2 , . . . , we get
c

An

c
Acn

=

n

c
n An .

c c

c

Acn

=

n

Acn .

=

n

n

Hence, Proposition 1.4(b) holds.
1.33 We have x ∈ B ∩
n An iff x ∈ B and x ∈
n An iff x ∈ B and x ∈ An for some n iff x ∈ B ∩ An
for some n iff x ∈ n (B ∩ An ). Hence, Proposition 1.5(a) holds. Applying that result to the sets B c
and Ac1 , Ac2 , . . . and using De Morgan’s laws, we get

B∪

An

= B

c c

c c

∪

n

c c
c

= B ∩

An

n
c

B c ∩ Acn

=
n

=

B c ∩ Acn

=

c

n

Bc

=

n

Acn

= B ∩

An

n

c
c

c

∪ Acn

c

n

(B ∪ An ).
n

Hence, Proposition 1.5(b) holds.
1.34
a) Referring first to Exercise 1.30(c) and then to Proposition 1.5(a) on page 14, we get
B=B∩

An

=

n

(B ∩ An ) =
n

(An ∩ B).
n

b) Let E be a subset of U . Then E ⊂ U = n An and, hence, from part (a), we have E =
c) Referring to Exercise 1.30(a), we get, for m = n,

n (An

∩ E).

(Am ∩ E) ∩ (An ∩ E) = Am ∩ E ∩ An ∩ E = (E ∩ E) ∩ (Am ∩ An )
= E ∩ (Am ∩ An ) = E ∩ ∅ = ∅.
Hence, A1 ∩ E, A2 ∩ E, . . . are pairwise disjoint.
d) Let E be a subset of U . Because A1 , A2 , . . . form a partition of U , we have n An = U . Hence, from
part (b), we have E = n (An ∩ E). And, because A1 , A2 , . . . form a partition of U , the sets A1 , A2 , . . .
are pairwise disjoint, which, by part (c), implies that A1 ∩ E, A2 ∩ E, . . . are pairwise disjoint. Therefore,
we see that E can be expressed as a disjoint union of the sets A1 ∩ E, A2 ∩ E, . . . .

Advanced Exercises
1.35
∞
∞
a) Let x ∈ ∞
k=n0 Ak , which means that x ∈ Ak
k=n Ak . Then there is an n0 ∈ N such that x ∈
n=1
for all k ≥ n0 . Now let n ∈ N and set m = max{n, n0 }. Then x ∈ Am ⊂ ∞
k=n Ak . Thus, we have shown
∞
∞
.
A
A
for
all
n
∈
N
,

which
means
that
x
∈
that x ∈ ∞
k=n k Consequently,
k=n k
n=1
∞

∞

∞

Ak
n=1 k=n

∞

⊂

Ak .
n=1 k=n


CHAPTER 1 Probability Basics

1-14

b) We see that x ∈ lim inf n→∞ An iff there is an n ∈ N such that x ∈ ∞
k=n Ak , which is the case
iff x ∈ Ak for all k ≥ n. Thus, lim inf n→∞ An consists of all elements of U that belong to all but a finite
number of the An s. Also, we see that x ∈ lim supn→∞ An iff for each n ∈ N , we have x ∈ ∞
k=n Ak ,
which is the case iff for each n ∈ N , there is a k ≥ n such that x ∈ Ak . Thus, lim supn→∞ An consists
of all elements of U that belong to an infinite number of the An s.
c) We first determine lim inf n→∞ An . To that end, we begin by noting that An ∩ An+1 = {0} for
all n ∈ N . Hence, as ∞
k=n Ak ⊂ An ∩ An+1 for all n ∈ N , we deduce that
∞
n→∞

∞

∞

lim inf An =

Ak

{0} = {0}.

⊂

n=1 k=n

n=1

However, from the definition of the An s, we see that 0 ∈ An for all n ∈ N and, consequently, it is clear

that {0} ⊂ lim inf n→∞ An . We have thus shown that lim inf n→∞ An = {0}.
Next we find lim supn→∞ An . We first note that A1 ⊃ A3 ⊃ A5 · · · and A2 ⊃ A4 ⊃ A6 · · ·. Therefore, ∞
k=n Ak = An ∪ An+1 for all n ∈ N and, hence,
∞

lim sup An =
n→∞

∞

∞

Ak
n=1 k=n

=

(An ∪ An+1 ).
n=1

Now, from the definition of the An s, we see that An ⊃ [−1, 0] for all odd n ∈ N and An ⊃ [0, 1] for
all even n ∈ N . Therefore, An ∪ An+1 ⊃ [−1, 1] for all n ∈ N and, hence, lim supn→∞ An ⊃ [−1, 1].
Conversely, suppose that x ∈
/ [−1, 1]. If x < −1, then there is an odd n ∈ N such that x < −1 − 1/n.
Thus, x ∈
/ An ∪ An+1 , which implies that x ∈
/ lim supn→∞ An . If x > 1, then there is an even n ∈ N
such that x > 1 + 1/n. Thus, x ∈
/ An ∪ An+1 , which implies that x ∈
/ lim supn→∞ An . Consequently,

no number outside of [−1, 1] is a member of lim supn→∞ An or, equivalently, lim supn→∞ An ⊂ [−1, 1].
We have thus shown that lim supn→∞ An = [−1, 1].
1.36 Define f : N → Z by
f (n) =

n/2,
−(n − 1)/2,

if n is even;
if n is odd,

We claim that f is one-to-one and onto Z, which will show that Z is countably infinite. Let z ∈ Z. If z is
a positive integer, then 2z is a even positive integer and, hence, f (2z) = 2z/2 = z. If z is a nonpositive
integer, then 1 − 2z is an odd positive integer and, hence, f (1 − 2z) = − (1 − 2z) − 1 /2 = z. Thus,
we have shown that f is onto Z. Next, we verify that f is one-to-one. Suppose then that f (m) = f (n).
Then either both m and n are even or both are odd. If they are both even, then we have m/2 = n/2 and,
hence, m = n; if they are both odd, then we have −(m − 1)/2 = −(n − 1)/2 and, hence, m = n.
1.37 Define f : N 2 → N by f (m, n) = 2m−1 (2n − 1). We claim that f is one-to-one and onto N ,
which will show that N 2 is countably infinite. Let k ∈ N . If k is even, let j denote the largest positive
integer such that 2j divides k. Then k = 2j (2n − 1) for some n ∈ N , and we have
f (j + 1, n) = 2(j +1)−1 (2n − 1) = 2j (2n − 1) = k.
If k is odd, then (k + 1)/2 ∈ N , and we have
f 1, (k + 1)/2 = 21−1 2 ·

k+1
− 1 = 1 · (k + 1) − 1 = k.
2

Thus, we have shown that f is onto N . Next, we establish that f is one-to-one. Suppose then
that f (m, n) = f (j, k), where, say, m ≥ j . Then, we have 2m−1 (2n − 1) = 2j −1 (2k − 1) or, equivalently, 2m−j (2n − 1) = 2k − 1. As 2k − 1 is odd, we must have m − j = 0, or m = j , in which case,

we have 2n − 1 = 2k − 1 and, hence, n = k. Therefore, (m, n) = (j, k).


1.2

Set Theory

1-15

1.38 For convenience, let us use the terminology 1-1 correspondence as an abbreviation for a function that
is one-to-one and onto. Suppose A is countable. Then, by definition, it is either finite or countably infinite.
If it is countably infinite, then, by definition, it is equivalent to N , which means there is a one-to-one and
onto function, f , from N to A. Letting sn = f (n), n ∈ N , we have that A is the range of the infinite
sequence {sn }∞
n=1 . If A is finite (and nonempty), then, by definition, there is an N ∈ N such that A is
equivalent to the first N positive integers. Let g be a one-to-one and onto function from {1, 2, . . . , N}
to A. Select x ∈ A and define sn = g(n) if n = 1, 2, . . . , N, and sn = x if n > N. Then A is the range
of the infinite sequence {sn }∞
n=1 .
Conversely, suppose A is the range of an infinite sequence, {sn }∞
n=1 . We claim that A is countable.
If A is finite, there is nothing to prove. So, assume that A is infinite. We will construct a 1-1 correspondence
from N to A, thereby proving that A is countably infinite and, hence, countable. Let n1 = 1. As A is
infinite, A \ {sn1 } = ∅. Therefore, because the range of {sn }∞
n=1 is A, the set { n ∈ N : sn = sn1 } is
not empty. Denote by n2 the smallest integer in that set. Note that n1 < n2 . Proceeding inductively,
note again that, because A is infinite, we have that A \ {sn1 , sn2 , . . . , snk } = ∅. Therefore, as the range
of {sn }∞
n=1 is A, the set { n ∈ N : sn = snj , 1 ≤ j ≤ k } is not empty. Denote by nk+1 the smallest
integer in that set and note that nk < nk+1 . We claim that the function f : N → A defined by f (k) = snk

is a 1-1 correspondence. By construction, f is one-to-one. So it remains to show that f is onto.
Let x ∈ A. Because the range of {sn }∞
n=1 is A, the set { n ∈ N : sn = x } is not empty. Let m be the
smallest integer in that set. If m = 1, then x = s1 = sn1 = f (1). Otherwise, let k be the smallest integer
such that m ≤ nk . As sn = x = sm for n < m, we have that sm = snj for 1 ≤ j ≤ k − 1, which implies
that m ≥ nk . Therefore, m = nk and, consequently, x = snk = f (k).
1.39 Let A be a countable set and f a function defined on A. If A = ∅, then f (A) = ∅ and, hence,
is countable. So, assume that A = ∅. We know from Exercise 1.38 that A is the range of an infinite
sequence, {sn }∞
n=1 . For each n ∈ N , define tn = f (sn ). Now, let y ∈ f (A). Then there is an x ∈ A
such that f (x) = y. Because A is the range of the infinite sequence {sn }∞
n=1 , there is an n ∈ N such
that sn = x. Therefore, y = f (x) = f (sn ) = tn . This result shows that f (A) is the range of the infinite
sequence {tn }∞
n=1 . Hence, by Exercise 1.38, f (A) is countable.
1.40 Let A be a countable set and let B ⊂ A. We claim that B is countable. If B = ∅, there is nothing to
prove; so, assume that B = ∅. This assumption implies that A is nonempty and, hence, by Exercise 1.38,
A is the range of an infinite sequence, {sn }∞
/ B.
n=1 . Choose x ∈ B. Let tn = sn if sn ∈ B, and tn = x if sn ∈
Then B is the range of the infinite sequence {tn }∞
.
Applying
Exercise
1.38
again,
we
conclude
that
B

n=1
is countable.
1.41 Let A and B be two countable sets. If either A or B is empty, then A × B = ∅ and, hence, is
countable. Therefore, let us assume that both A and B are nonempty. By Exercise 1.38, each of A and B
∞
2
is the range of an infinite sequence, say, {an }∞
n=1 and {bn }n=1 , respectively. Define f : N → A × B
by f (m, n) = (am , bn ). From Exercise 1.37, we know that N 2 is countable and, clearly, f is onto A × B.
Thus, A × B is the image of N 2 under f . Applying Exercise 1.39, we conclude that A × B is countable.
We next use mathematical induction to prove that the Cartesian product of a finite number of countable
sets is countable. We have just verified that result for n = 2. Assuming its truth for n − 1, we now prove
it for n. Thus, let A1 , . . . , An be countable sets. For convenience, set Bk = ×jk=1 Aj for each k ∈ N .
From the induction assumption, we know that Bn−1 is countable and, hence, so is Bn−1 × An . Let us
define f : Bn−1 × An → Bn by f (a1 , . . . , an−1 ), an = (a1 , . . . , an−1 , an ). Clearly, f is one-to-one
and onto and, hence, as Bn−1 × An is countable, so is Bn = ×jn=1 Aj .
1.42 We know from Exercise 1.36 that Z is countable. Hence, by Exercise 1.41, so is Z × N . Define f : Z × N → Q by f (z, n) = z/n. Clearly, f is onto Q, and, hence, Q is the image of Z × N
under f . Therefore, we conclude from Exercise 1.39 that Q is countable.


1-16

CHAPTER 1 Probability Basics

1.43
a) Suppose to the contrary that [0, 1) is countable. Let {xn }∞
n=1 be an enumeration of its elements,
that is, the function f : N → [0, 1) defined by f (n) = xn is one-to-one and onto. For each n ∈ N ,
let 0.dn1 dn2 . . . denote the unique decimal expansion of xn not containing only finitely many digits
−n = 0.y y . . ., where y = 1 if d

differing from 9. Now consider the number y = ∞
1 2
n
nn = 0,
n=1 yn 10
and yn = 0 otherwise. Note that y ∈ [0, 1) and, hence, as {xn }∞
is
an
enumeration
of
[0,
1),
we
must
n=1
have y = xn for some n ∈ N . However, for each n ∈ N , we have y = xn because the decimal expansions
of y and xn differ at the nth decimal digit and neither decimal expansion contains only finitely many
digits differing from 9. Thus, we have a contradiction. Therefore, [0, 1) is uncountable.
b) Suppose to the contrary that (0, 1) is countable. Let {xn }∞
n=1 be an enumeration of its elements.
Define f : N → [0, 1) by f (1) = 0 and f (n) = xn−1 if n ≥ 2. Then f is a one-to-one function from N
onto [0, 1), which implies that [0, 1) is countable, a contradiction to part (a).
c) Define f : (0, 1) → (a, b) by f (x) = a + (b − a)x. Now, let y ∈ (a, b). Upon solving the equation y = a + (b − a)x for x, we get the unique number x = (y − a)/(b − a), which, clearly, is in (0, 1).
Consequently, f is a one-to-one function from (0, 1) onto (a, b), which, by part (b), implies that (a, b)
is uncountable.
d) From Exercise 1.40, we know that any subset of a countable set is countable or, equivalently, if a
set contains an uncountable subset, then it must be uncountable. Because any nondegenerate interval
contains a bounded interval of the form (a, b), it follows from part (c) that any nondegenerate interval is
uncountable.
1.44 Denote the countable collection of sets by C. If C is empty, then its union is empty and, hence,

countable. So, we can assume that C is nonempty. Without loss of generality, we can also assume that
each member of C is nonempty; otherwise, we simply discard all empty members of C, which does not
affect the union. Because C is nonempty and countable, Exercise 1.38 shows that it is the range of an
∞
infinite sequence, say, {An }∞
n=1 An is countable. Because each member
n=1 . We need to prove that A =
(n) ∞
of C is countable, Exercise 1.38 shows that each An is the range of an infinite sequence, say, xm m=1 .
(n)

Now, define f : N 2 → A by f (m, n) = xm . We note that f is onto. Furthermore, by Exercise 1.37, we
know that N 2 is countable. Therefore, in view of Exercise 1.39, we conclude that A is countable, being
the image of N 2 under f .
1.45
a) Suppose that I is finite, say, I = {1, 2, . . . , N}. On the one hand, by Definition 1.5, ×N
n=1 An is
the set of all ordered N-tuples (a1 , a2 , . . . , aN ), where an ∈ An for 1 ≤ n ≤ N . On the other hand,
by the definition of the Cartesian product of an indexed collection, ×N
n=1 An is the set of all functions x on {1, 2, . . . , N} such that x(n) ∈ An for 1 ≤ n ≤ N . Identifying each such function x with
the ordered N -tuple x(1), x(2), . . . , x(N ) , we obtain a one-to-one correspondence between the members of the Cartesian product ×N
n=1 AN as defined by Definition 1.5 and the members of the Cartesian
N
product ×n=1 An of the indexed collection {Aι }ι∈I .
b) False, it is not necessarily the case that the Cartesian product of a countable number of countable sets
is countable. For instance, let D denote the set of decimal digits and let E = D ∞ = D × D × D × · · ·,
the Cartesian product of a countable number of countable (actually, finite) sets. Define f : E → [0, 1]
−n
by f (d1 , d2 , . . .) = ∞
n=1 dn 10 . We have f (E) = [0, 1]; that is, [0, 1] is the image of f under E.

If E were countable, then, by Exercise 1.39, so would be [0, 1]. However, by Exercise 1.43(d), we know
that [0, 1] is uncountable. Hence, E must be uncountable.


Review Exercises for Chapter 1

1-17

Review Exercises for Chapter 1
Basic Exercises
1.46
a) Referring to Table 1.1 on page 4, we see that 16 of the states are in the South and, of those, 11 seceded
from the Union. Therefore, the probability that a randomly selected state in the South seceded from the
union is 11/16, or 0.6875.
b) The population under consideration consists of the 16 states in the South.
1.47 From the table, the total number of winners for the years 1901–1997 is 190 + 71 + · · · + 87 = 448.
a) The number of winners from Japan is 4. Hence, the probability that the recipient selected is from
Japan is 4/448 ≈ 0.00893.
b) The number of winners from either France or Germany is 25 + 61 = 86. Hence, the probability that
the recipient selected is from either France or Germany is 86/448 ≈ 0.192.
c) The number of winners from any country other than the United States is 448 − 190 = 258. Hence, the
probability that the recipient selected is from any country other than the United States is 258/448 ≈ 0.576.
1.48
a) Referring to the frequentist interpretation of probability on page 5, we see that, in a large number of
tosses of the die, the result will be 3 about 1/6 of the time.
b) Again referring to the frequentist interpretation of probability, we see that, in a large number of tosses
of the die, the result will be 3 or more about 2/3 of the time.
c) From part (a), we see that, in 10,000 tosses, the die will come up 3 about 10,000 · (1/6) times, or
roughly 1667 times.
d) From part (b), we see that, in 10,000 tosses, the die will come up 3 or more about 10,000 · (2/3) times,

or roughly 6667 times.
1.49 Consider the random experiment of selecting one voter (at random) from the population and let E
denote the event that the voter chosen will vote yes on the proposition. Because 60% of the voters will
vote yes, we know that P (E) = 0.6. Now, choosing n voters at random with replacement is equivalent
to independently repeating the random experiment n times. Hence, from the frequentist interpretation of
probability, specifically Relation (1.1) on page 5, we have, for large n, that
n(Y ) = n(E) ≈ P (E) · n = 0.6n.
1.50
a) Squaring each element of the set S = {−2, −1, 0, 1, 2}, we find that the elements of { x 2 : x ∈ S }
are 0, 1, and 4.
b) If −2 < x < 2, then 0 ≤ x 2 < 4 and, vice-versa. Hence, { x 2 : −2 < x < 2 } = [0, 4).
1.51 The eight members of {0, 1}3 are (0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 0), (1, 0, 1), (1, 1, 0),
and (1, 1, 1). By associating 0 with a tail and 1 with a head, we can regard the eight members of {0, 1}3
as the eight possible outcomes of the experiment of tossing a coin three times. We now consider the
eight members of {0, 1}3 a finite population. If we select a member at random from that population, then
each member is equally likely to be the one obtained. Hence, that random experiment can be regarded as
tossing a balanced coin three times. Note that there are three members of the population corresponding
to getting two heads and one tail, namely, (0, 1, 1), (1, 0, 1), and (1, 1, 0). Therefore, the probability that
you get two heads and one tail when you toss a balanced coin three times equals the probability that a
member selected at random from {0, 1}3 consists of exactly two 1s and one 0, which is 3/8, or 0.375.


CHAPTER 1 Probability Basics

1-18

1.52 We can proceed in several ways to obtain the simplification. One way is to apply the distributive
law of Proposition 1.2(a) on page 12, as follows:
(A ∩ B) ∪ (A ∩ B c ) = A ∩ (B ∪ B c ) = A ∩ U = A.
1.53 For each n ∈ N , let An = 1/(n + 1), 1/n and Bn = 1/(n + 1), 1/n .

a) Set A = ∞
n=1 An . If x ∈ A, then there is an n ∈ N such that x ∈ An and, so, 1/(n + 1) < x ≤ 1/n].
But then we have
1
1
< x ≤ ≤ 1.
0<
n+1
n
Thus, A ⊂ (0, 1]. Conversely, suppose that x ∈ (0, 1]. Then x > 0. Let n be the smallest positive integer
such that 1/(n + 1) < x. Then 1/(n + 1) < x ≤ 1/n so that x ∈ An , which, in turn, implies that x ∈ A.
Thus, (0, 1] ⊂ A. We have shown that A ⊂ (0, 1] and (0, 1] ⊂ A. Consequently, A = (0, 1].
b) Set B = ∞
n=1 Bn . If x ∈ B, then there is an n ∈ N such that x ∈ Bn and, so, 1/(n + 1) ≤ x ≤ 1/n].
But then we have
1
1
≤ x ≤ ≤ 1.
0<
n+1
n
Thus, B ⊂ (0, 1]. Conversely, suppose that x ∈ (0, 1]. Then x > 0. Let n be the smallest positive integer
such that 1/(n + 1) < x. Then 1/(n + 1) < x ≤ 1/n so that x ∈ Bn , which, in turn, implies that x ∈ B.
Thus, (0, 1] ⊂ B. We have shown that B ⊂ (0, 1] and (0, 1] ⊂ B. Consequently, B = (0, 1].
c) The sets in the union in part (a) are pairwise disjoint. Indeed, let m = n, say, m < n. Then, if x ∈ An ,
we have x ≤ 1/n ≤ 1/(m + 1) and, hence, x ∈
/ Am . Consequently, An ∩ Am = ∅. The sets in the union
in part (b) are not pairwise disjoint. For instance, we have
B1 ∩ B2 = [1/2, 1] ∩ [1/3, 1/2] = {1/2} = ∅.
However,

∞

Bn ⊂ B1 ∩ B3 = [1/2, 1] ∩ [1/4, 1/3] = ∅.
n=1

Consequently,

∞
n=1 Bn

= ∅.

1.54 You may find it helpful to refer to the discussion of finite and infinite sets on pages 19 and 20. Also,
note that other methods can be used to obtain the following results.
a) We have
{1, 2, 3} × {2, 3, 4} = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4)}.
Thus, {1, 2, 3} × {2, 3, 4} consists of nine elements and, hence, is finite.
b) Let A = {1, 2, 3} × [2, 4]. We have
A = (1, y) : y ∈ [2, 4] ∪ (2, y) : y ∈ [2, 4] ∪ (3, y) : y ∈ [2, 4] .
Denote by B the first set in the union of the preceding display. Define f : [2, 4] → B by f (y) = (1, y).
Clearly, f is one-to-one and onto. Hence, from Exercise 1.43, we deduce that B is uncountable.
As B ⊂ A, it follows from Exercise 1.40 that A is uncountable.
c) Let A be as in part (b) and set C = [2, 4] × {1, 2, 3}. Define f : A → C by f (x, y) = (y, x). Clearly, f
is one-to-one and onto. From part (b), we know that A is uncountable; hence, so is C.
d) Let D = {1, 2, 3} × {1, 2, 3, . . .}. From Exercise 1.41, we see that D is countable, being the Cartesian
product of two countable sets (one finite and the other countably infinite). Obviously, D is infinite and,
therefore, we conclude that it is countably infinite.


Review Exercises for Chapter 1

1-19

e) Let E = [1, 3] × [2, 4] and let B be as in part (b). As we have seen, B is uncountable. Therefore,
because E ⊃ B, we deduce from Exercise 1.40 that E is uncountable.
f) We note that ∞
n=1 {n, n + 1, n + 2} = N , which is countably infinite.
g) Let An = {n, n + 1, n + 2} and F = ∞
n=1 An . We have
F ⊂ A1 ∩ A4 = {1, 2, 3} ∩ {4, 5, 6} = ∅.
Therefore, F = ∅ and, hence, is finite.
1.55
a) From De Morgan’s laws,
{3, 4}c ∩ {4, 5}c

c

= {3, 4}c

c

∪ {4, 5}c

c

= {3, 4} ∪ {4, 5} = {3, 4, 5}.

b) We have
{3, 4}c ∩ {4, 5}c

c

= {. . . , −2, −1, 0, 1, 2, 5, 6, . . .} ∩ {. . . , −2, −1, 0, 1, 2, 3, 6, 7, . . .}
= {. . . , −2, −1, 0, 1, 2, 6, 7, . . .}

c

c

= {3, 4, 5}.

Theory Exercises
1.56 Let B1 = A1 and, for k ≥ 2, set
k−1

Bk = Ak ∩

c

Aj

= Ac1 ∩ · · · ∩ Ack−1 ∩ Ak .

j =1

Observe that, for all k ∈ N , we have Bk ⊂ Ak and Bk ⊂ Ajc for 1 ≤ j ≤ k − 1. Let m = n, say, m < n.
Then Bm ⊂ Am and Bn ⊂ Acm , so that
Bm ∩ Bn ⊂ Am ∩ Acm = ∅.
Consequently, Bm ∩ Bn = ∅, and we see that B1 , B2 , . . . are pairwise disjoint sets. We use these sets
in parts (a) and (b).

a) For convenience, set A(n) = jn=1 Aj and B (n) = jn=1 Bj . As Bj ⊂ Aj for all j ∈ {1, 2, . . . , n}, we
have B (n) ⊂ A(n) . Conversely, suppose that x ∈ A(n) . Then x ∈ Aj for some j ∈ {1, 2, . . . , n}. Let k be
the smallest such j . Then x ∈ Ak and x ∈
/ Aj for 1 ≤ j ≤ k − 1, which means that x ∈ Bk and, consequently, that x ∈ B (n) . Therefore, A(n) ⊂ B (n) . We have now shown that B (n) ⊂ A(n) and A(n) ⊂ B (n) .
Hence, A(n) = B (n) .
b) We use the notation of part (a) and recall that A(n) = B (n) for all n ∈ N . Let A = ∞
n=1 An
∞
∞
∞
(n)
(n)
and B = n=1 Bn . We observe that A = n=1 A and B = n=1 B . Hence,
∞

∞

B (n) = B.

A(n) =

A=
n=1

n=1

1.57
a) From the distributive law of Proposition 1.2(a) on page 12 and the fact that A ∩ C ⊂ A, we get
(A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C) ⊂ A ∪ (B ∩ C).
b) Answers will vary. Note, however, from the solution to part (a) that any choice of A, B, and C in

which A ∩ C = A will do the trick. For instance, take A = {1} and B = C = ∅. Then
(A ∪ B) ∩ C = ({1} ∪ ∅) ∩ ∅ = ∅ = {1} = {1} ∪ ∅ = {1} ∪ (∅ ∩ ∅) = A ∪ (B ∩ C).


1-20

CHAPTER 1 Probability Basics

c) If A ⊂ C, then A ∩ C = A. Hence, from the distributive law of Proposition 1.2(a),
(A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C) = A ∪ (B ∩ C).
d) If (A ∪ B) ∩ C = A ∪ (B ∩ C), then
A ⊂ A ∪ (B ∩ C) = (A ∪ B) ∩ C ⊂ C.
1.58
a) Let C = { C ⊂ U : A ⊂ C and B ⊂ C }. We want to prove that A ∪ B = C∈C C. To begin, we note
that, because A ⊂ A ∪ B and B ⊂ A ∪ B, we have A ∪ B ∈ C. Thus, A ∪ B ⊃ C∈C C. Conversely,
let C ∈ C. Then A ⊂ C and B ⊂ C, which implies that A ∪ B ⊂ C. Therefore, A ∪ B ⊂ C∈C C.
b) If A ⊂ D and B ⊂ D, then A ∪ B ⊂ D. Hence, we see that A ∪ B is the smallest set that contains
both A and B as subsets.
c) Let D = { C ⊂ U : A ⊃ C and B ⊃ C }. We want to prove that A ∩ B = C∈D C. To begin, we note
that, because A ⊃ A ∩ B and B ⊃ A ∩ B, we have A ∩ B ∈ D. Thus, A ∩ B ⊂ C∈D C. Conversely,
let C ∈ D. Then A ⊃ C and B ⊃ C, which implies that A ∩ B ⊃ C. Therefore, A ∩ B ⊃ C∈D C.
d) If A ⊃ D and B ⊃ D, then A ∩ B ⊃ D. Hence, we see that A ∩ B is the largest set that is contained
as a subset by both A and B.
e) Let A1 , A2 , . . . be a countable sequence of subsets of U .
Generalization of part (a): Let E = { C ⊂ U : An ⊂ C for all n }. We want to prove that
An =
n

C.
C∈E

For convenience, set A = n An . Because An ⊂ A for all n, we have A ∈ E. Thus, A ⊃ C∈E C.
Conversely, let C ∈ E. Then An ⊂ C for all n, which implies that A ⊂ C. Therefore, A ⊂ C∈E C.
Generalization of part (b): If An ⊂ D for all n, then
smallest set that contains all An s as subsets.

n An

⊂ D. Hence, we see that

n An

is the

Generalization of part (c): Let F = { C ⊂ U : An ⊃ C for all n }. We want to prove that
An =
n

C.
C∈F

For convenience, set A = n An . Because An ⊃ A for all n, we have A ∈ F. Thus, A ⊂ C∈F C.
Conversely, let C ∈ F. Then An ⊃ C for all n, which implies that A ⊃ C. Therefore, A ⊃ C∈F C.
Generalization of part (d): If An ⊃ D for all n, then
largest set that is contained as a subset of all An s.

n An

⊃ D. Hence, we see that

n An

is the

1.59
a) This result is true. Indeed, we have (x, y) ∈ A × (B ∪ C) iff x ∈ A and y ∈ B ∪ C iff x ∈ A
and either y ∈ B or y ∈ C iff either x ∈ A and y ∈ B or x ∈ A and y ∈ C iff either (x, y) ∈ A × B
or (x, y) ∈ A × C iff (x, y) ∈ (A × B) ∪ (A × C). Hence, A × (B ∪ C) = (A × B) ∪ (A × C).
b) This result is true. Indeed, we have (x, y) ∈ A × (B ∩ C) iff x ∈ A and y ∈ B ∩ C iff x ∈ A
and y ∈ B and y ∈ C iff x ∈ A and y ∈ B and x ∈ A and y ∈ C iff (x, y) ∈ A × B and (x, y) ∈ A × C
iff (x, y) ∈ (A × B) ∩ (A × C). Hence, A × (B ∩ C) = (A × B) ∩ (A × C).
c) This result is not always true. For instance, let A = {0} and B = {1}. Then
A × B = {(0, 1)} = {(1, 0)} = B × A.


Review Exercises for Chapter 1

1-21

Advanced Exercises
1.60 Refer to the table in the problem statement.
a) The number of left-handed individuals is 71 + 92 = 163. Hence, the probability that the person
obtained is left-handed is 163/525 ≈ 0.310.
b) The number of colorblind individuals who are not ambidextrous is 71 + 61 = 132. Hence, the probability that the person obtained is colorblind but not ambidextrous is 132/525 ≈ 0.251.
c) The number of left-handed individuals is 71 + 92 = 163 of which 71 are colorblind. Hence, if the
person selected is left-handed, the probability that he or she is colorblind is 71/163 ≈ 0.436.
d) The number of colorblind individuals is 71 + 61 + 37 = 169 of which 71 are left-handed. Hence, if
the person selected is colorblind, the probability that he or she is left-handed is 71/169 ≈ 0.420.
e) The number of left-handed, colorblind women is 24. Hence, the probability that the person obtained
is a left-handed, colorblind woman is 24/525 ≈ 0.0457.

1.61 The probability is 0.75 that a randomly selected adult female believes that having a "cyber affair" is
cheating. Therefore, the odds against an adult female believing that having a "cyber affair" is cheating are
1 − 0.75 to 0.75

or

0.25 to 0.75

or

1 to 3.

1.62
a) We have
/ B } = { x : x ∈ U and x ∈
/ B } = U \ B.
Bc = { x : x ∈
b) We have x ∈ A \ B iff x ∈ A and x ∈
/ B iff x ∈ A and x ∈ B c iff x ∈ A ∩ B c . Consequently, we have
c
shown that A \ B = A ∩ B .
c) Applying part (b) and De Morgan’s law of Proposition 1.1(b) on page 11, we get
(A \ B)c = A ∩ B c

c

= Ac ∪ B c

c

= Ac ∪ B.

1.63 We note that, by definition, x ∈ A △ B iff either x ∈ A or x ∈ B, and x ∈
/ A ∩ B, which is the
c
c
case iff x ∈ A ∪ B and x ∈ (A ∩ B) iff x ∈ (A ∪ B) ∩ (A ∩ B) . Thus, A △ B = (A ∪ B) ∩ (A ∩ B)c .
Applying De Morgan’s laws, the distributive laws, and Exercise 1.62(b), we get
A △ B = (A ∪ B) ∩ (A ∩ B)c = (A ∪ B) ∩ Ac ∪ B c = (A ∪ B) ∩ Ac ∪ (A ∪ B) ∩ B c
= (A ∩ Ac ) ∪ (B ∩ Ac ) ∪ (A ∩ B c ) ∪ (B ∩ B c ) = ∅ ∪ (B \ A) ∪ (A \ B) ∪ ∅
= (A \ B) ∪ (B \ A).
1.64
a) We note that x ∈ B △ C iff x is a member of exactly one of B and C. Hence, x ∈
/ B △ C iff either x
is in both B and C or x is in neither B nor C. In other words,
(B △ C)c = (B ∩ C) ∪ (B c ∩ C c ),
a result that we could also obtain by applying properties of set operations and results from Exercises 1.62
and 1.63. Referring now to those two exercises, we get
A △ (B △ C) = A ∩ (B △ C)c ∪ (B △ C) ∩ Ac
= A ∩ (B ∩ C) ∪ (B c ∩ C c )

∪

(B ∩ C c ) ∪ (C ∩ B c ) ∩ Ac

= A ∩ B ∩ C ∪ A ∩ B c ∩ C c ∪ B ∩ C c ∩ Ac ∪ C ∩ B c ∩ Ac .
Replacing A by C, B by A, and C by B in the previous display, we find that C △ (A △ B) = A △ (B △ C).
The required result now follows from the easily established fact that E △ F = F △ E.

CHAPTER 1 Probability Basics

1-22

b) We have
A △ U = A ∩ U c ∪ U ∩ Ac = (A ∩ ∅) ∪ Ac = ∅ ∪ Ac = Ac .
c) We have
A △ ∅ = A ∩ ∅c ∪ ∅ ∩ Ac = (A ∩ U ) ∪ ∅ = A ∪ ∅ = A.
d) We have
A △ A = A ∩ Ac ∪ A ∩ Ac = ∅ ∪ ∅ = ∅.
1.65
a) Applying properties of set operations, we get
A ∩ (B △ C) = A ∩ (B ∩ C c ) ∪ (C ∩ B c ) = A ∩ (B ∩ C c ) ∪ A ∩ (C ∩ B c )
= A ∩ B ∩ C c ∪ A ∩ C ∩ B c = (A ∩ B) ∩ (Ac ∪ C c ) ∪ (A ∩ C) ∩ (Ac ∪ B c )
= (A ∩ B) ∩ (A ∩ C)c ∪ (A ∩ C) ∩ (A ∩ B)c = (A ∩ B) △ (A ∩ C).
b) Again applying properties of set operations, we get
(A ∪ B) △ (A ∪ C) = (A ∪ B) ∩ (A ∪ C)c ∪ (A ∪ C) ∩ (A ∪ B)c
= (A ∪ B) ∩ (Ac ∩ C c ) ∪ (A ∪ C) ∩ (Ac ∩ B c )
= B ∩ Ac ∩ C c ∪ C ∩ Ac ∩ B c = Ac ∩ (B ∩ C c ) ∪ (C ∩ B c )
= Ac ∩ (B △ C).
Therefore,
(A ∪ B) △ (A ∪ C) = Ac ∩ (B △ C) ⊂ B △ C ⊂ A ∪ (B △ C).
Thus, we have shown that (A ∪ B) △ (A ∪ C) ⊂ A ∪ (B △ C).
c) Suppose that A = ∅. Then
A ∪ (B △ C) = ∅ ∪ (B △ C) = B △ C = (∅ ∪ B) △ (∅ ∪ C) = (A ∪ B) △ (A ∪ C).
Conversely, suppose that A ∪ (B △ C) = (A ∪ B) △ (A ∪ C). Then, referring to the solution to part (b),
we find that
A ⊂ A ∪ (B △ C) = (A ∪ B) △ (A ∪ C) = Ac ∩ (B △ C) ⊂ Ac .
From this result, we conclude that A = A ∩ A ⊂ A ∩ Ac = ∅; that is, A = ∅. We have therefore shown
that A ∪ (B △ C) = (A ∪ B) △ (A ∪ C) precisely when A = ∅.

1.66 Suppose that B = ∅. Then, from Exercise 1.64(c), we have A △ B = A △ ∅ = A. Conversely,
suppose that A = A △ B. Then, applying in turn parts (c), (d), (a), and (d) of Exercise 1.64, we find that
B = B △ ∅ = ∅ △ B = (A △ A) △ B = A △ (A △ B) = A △ A = ∅.